Question

A small sphere of mass m = 6.00 g and charge q1 = 32.1 nC is attached to the end of a string and hangs vertically as in the figure. A second charge of equal mass and charge q2 = -58.0 nC is located below the first charge a distance d = 2.00 cm below the first charge as in the figure. M

(a) Find the tension in the string.

(b) If the string can withstand a maximum tension of 0, 180 N, what is the smallest value d can have before the string breaks?  

Answer 1

a) The mass is being pulled down with 2 force, gravity and electric attraction, and pulled up with tension of rope. Since it is not moving, we can say up force equals down force, or:

T = mg + Fe

To find Fe, the force of electricity, we use coloumbs law:

Fe = kq1q2/r2
Fe = (9*10^9)(32.1*10^-9)(58*10^-9) / (.02)^2
Fe = .0418 N

So we put it into the first equation:

T = mg + Fe
T = ( .006)(9.8) + .0418 N
T = 0.1006 N

b) We, we know T and mg, but Fe unknown, so:

T = mg + Fe
.180 N = (.006)(9.8) + Fe
Fe = .1212 N

We now plug this into coloumbs law and solve for r:

Fe = kq1q2/r2
.1212 N = (9*10^9)(32.1*10^-9)(58*10^-9) / (r)2
r = .01174 m or 1.174 cm